 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/30k/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

n < N

A_{n + 1} - A_{n} = V_{n} - E [V_{n + 1} ∣ F_{n}] \geq 0.

V_{n} = max {Z_{n}, E [V_{n + 1} ∣ F_{n}]}

shows that if

V_{n} > Z_{n}

, then

V_{n} = E [V_{n + 1} ∣ F_{n}]

and the increment of

A

is zero. Conversely, if that increment is positive, the maximum must be attained by

Z_{n}

, so

V_{n} = Z_{n}

. The two nonnegative quantities therefore satisfy

min {V_{n} - Z_{n}, A_{n + 1} - A_{n}} = 0.

For

n = N

, this also holds because

V_{N} = Z_{N}

, regardless of the convention

A_{N + 1} = \infty

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